Class 9 I'm Up and Down, and Round and Round Mock Test (CBSE)

Topics in I'm Up and Down, and Round and Round

• Definitions: centre, radius, chord, arc
• Symmetries of a circle
• Circles through given points
• Chords and the angles they subtend
• Perpendicular bisector of a chord
• Distance of chords from the centre
• Angles subtended by an arc
• Concyclic points

I'm Up and Down, and Round and Round practice questions with answers

1. Q1Multiple choice·1 mark
   A round kitchen table has centre O. A chord AB on its top subtends an angle of $120^\circ$ at O. A point P lies on the major arc AB of the table's edge. What is $\angle APB$?

   1. (a)$120^\circ$
   2. (b)$240^\circ$
   3. (c)$60^\circ$
   4. (d)$30^\circ$
   Show answer and step marks

   1. Correct option: (c) $60^\circ$. The angle subtended by an arc at the centre is twice the angle it subtends at any point on the remaining part of the circle, so $\angle APB = \frac{1}{2} \times 120^\circ = 60^\circ$.+1

   Answer: (c)
2. Q2Multiple choice·1 mark
   Which statement about the symmetry of a circle is correct?

   1. (a)It has exactly two lines of symmetry.
   2. (b)It has exactly four lines of symmetry.
   3. (c)It has infinitely many lines of symmetry, each passing through its centre.
   4. (d)It has no line of symmetry because it has no corners.
   Show answer and step marks

   1. Correct option: (c) It has infinitely many lines of symmetry, each passing through its centre.+1

   Answer: (c)
3. Q3Multiple choice·1 mark
   A chord of a circle is equal to its radius. The angle subtended by this chord at a point on the major arc is:

   1. (a)$30^\circ$
   2. (b)$60^\circ$
   3. (c)$90^\circ$
   4. (d)$45^\circ$
   Show answer and step marks

   1. Correct option: (a) $30^\circ$. The triangle formed by the chord and the radii to its ends is equilateral, so the angle subtended at the centre is $60^\circ$. The angle subtended by the arc at any point on the remaining part of the circle is half of the angle at the centre, which is $30^\circ$.+1

   Answer: (a)
4. Q4Short answer·2 marks
   Points $A$, $B$ and $C$ do not lie on one straight line. Explain how the centre of the circle passing through $A$, $B$ and $C$ can be located.
   Show answer and step marks

   1. The centre is equidistant from $A$ and $B$, so it lies on the perpendicular bisector of $AB$.+1
   2. The centre is also equidistant from $B$ and $C$, so it lies on the perpendicular bisector of $BC$; the centre is the point where these two perpendicular bisectors meet.+1

   Answer: Draw the perpendicular bisectors of $AB$ and $BC$; their point of intersection is the centre of the circle.
5. Q5Short answer·3 marks
   In a circle with centre $O$, chords $AB$ and $CD$ are equal in length. The angle subtended by chord $AB$ at the centre is $(3x+10)^\circ$ and the angle subtended by chord $CD$ at the centre is $(5x-20)^\circ$. Find the measure of $\angle AOB$.
   Show answer and step marks

   1. Equal chords of a circle subtend equal angles at the centre, so $\angle AOB = \angle COD$.+1
   2. Equate the two angles: $3x+10 = 5x-20$.+1
   3. Solve: $2x = 30$, so $x = 15$, and $\angle AOB = 3(15)+10 = 55^\circ$.+1

   Answer: $\angle AOB = 55^\circ$
6. Q6Short answer·3 marks
   $ABCD$ and $PQRS$ are convex quadrilaterals. In $ABCD$, $\angle A = 72^\circ$ and $\angle C = 108^\circ$. In $PQRS$, $\angle P = 65^\circ$ and $\angle R = 110^\circ$. Using these angles, decide which quadrilateral has all four of its vertices lying on a circle and which does not. Justify your answer.
   Show answer and step marks

   1. For $ABCD$, $\angle A + \angle C = 72^\circ + 108^\circ = 180^\circ$, a pair of opposite angles adding to $180^\circ$.+1
   2. Since the opposite angles of $ABCD$ are supplementary, its vertices $A$, $B$, $C$, $D$ are concyclic.+1
   3. For $PQRS$, $\angle P + \angle R = 65^\circ + 110^\circ = 175^\circ \neq 180^\circ$, so its opposite angles are not supplementary and $PQRS$ is not cyclic. Hence $ABCD$ lies on a circle and $PQRS$ does not.+1

   Answer: $ABCD$ is cyclic (its vertices are concyclic) because $\angle A + \angle C = 180^\circ$; $PQRS$ is not cyclic because $\angle P + \angle R = 175^\circ \neq 180^\circ$.

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